38B. Lookup and Insertion in Binary Search Trees

Lookup

---

Functions can take advantage of the ordering requirement in a way that is similar to what binary search does. A membership testing function can exploit the following facts.

1. An empty tree has no members.

2. If the root of T contains x, then tree T clearly contains x.

3. If the root of T contains k and x < k then, if x occurs in T at all, it must occur in T 's left subtree, by the ordering requirement for binary search trees.

4. Similarly, if the root of T contains k and x > k, then x occurs in T if and only if x occurs in T 's right subtree.

  //==========================================
  //               member
  //==========================================
  // member(x,T) returns true if x is in binary
  // search tree T.
  //==========================================

  bool member(const int x, const Tree> T)
  {
    if(T == NULL)
    {
      return false;
    }
    else if(x == T->item)
    {
      return true;
    }
    else if(x < T->item)
    {
      return member(x, T->left);
    }
    else
    {
      return member(x, T->right);
    }
  }

Notice the similarity between that and binary search.

Insertion

---

To insert a value, just find where that value would have been, had it already been in the tree, then add the value as a new leaf. For example, inserting 13 as shown below would result in the following change.

The actual insertion takes place when a null tree is encountered. The null tree is replaced by a leaf. An immediate consequence is:

If the value x to be inserted is already in the tree, nothing is done.

  //==========================================
  //                insert
  //==========================================
  // insert(x,T) inserts x (destructively) into
  // binary search tree T. If x is already a
  // member of T, insert does nothing.
  //==========================================

  void insert(int x, Tree& T)
  {
    if(T == NULL)
    {
      T = new Node(x, NULL, NULL);
    }
    else if(x < T->item)
    {
      insert(x, T->left);
    }
    else if(x > T->item)
    {
      insert(x, T->right);
    }
  }

Notice that parameter T is passed by reference. The first case needs to change the pointer that is stored in T. Also notice that there is no case for x = T->item since insert is supposed to do nothing when x already occurs in the tree.

Exercises

---

1. What tree do you get if you insert 25 into the following binary search tree?

   
   Answer

2. What tree do you get if you insert 7 into the following binary search tree?

   
   Answer

3. What tree do you get if you insert 11 into the following binary search tree?

   
   Answer